A particle performs SHM with a period `T` and amplitude a. The mean velocity of particle over the time interval during which it travels `a//2` from the extreme position is

Options

A. `6a//T`
B. `2 a//T`
C. `3 a//T`
D. `a//2T`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the setup of the problem
The particle is performing Simple Harmonic Motion (SHM) with a given amplitude \( a \) and period \( T \). The particle starts from the extreme position and travels a distance of \( \frac{a}{2} \) towards the mean position.

### Step 2: Define the displacement in SHM
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x = A \cos(\omega t) \]
where \( A \) is the amplitude, and \( \omega \) is the angular frequency given by:
\[ \omega = \frac{2\pi}{T} \]

### Step 3: Set the initial and final positions
- The initial position (extreme position) is at \( x = a \).
- The final position (after traveling \( \frac{a}{2} \)) is:
\[ x = a - \frac{a}{2} = \frac{a}{2} \]

### Step 4: Calculate the time taken to travel from the extreme position to \( \frac{a}{2} \)
Using the displacement equation:
\[ \frac{a}{2} = A \cos(\omega t) \]
Substituting \( A = a \):
\[ \frac{a}{2} = a \cos(\omega t) \]
Dividing both sides by \( a \):
\[ \frac{1}{2} = \cos(\omega t) \]
Taking the inverse cosine:
\[ \omega t = \cos^{-1}\left(\frac{1}{2}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ \omega t = \frac{\pi}{3} \]

### Step 5: Solve for time \( t \)
Substituting \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{3} \]
Solving for \( t \):
\[ t = \frac{\pi}{3} \cdot \frac{T}{2\pi} = \frac{T}{6} \]

### Step 6: Calculate the mean velocity
The mean velocity \( v_{mean} \) is defined as the total displacement divided by the total time taken. The total displacement is \( \frac{a}{2} \) and the total time taken is \( \frac{T}{6} \):
\[ v_{mean} = \frac{\text{Displacement}}{\text{Time}} = \frac{\frac{a}{2}}{\frac{T}{6}} = \frac{a}{2} \cdot \frac{6}{T} = \frac{3a}{T} \]

### Final Answer
The mean velocity of the particle over the time interval during which it travels \( \frac{a}{2} \) from the extreme position is:
\[ \boxed{\frac{3a}{T}} \]